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I am stuck on the following problems:

  1. How can I show that the set of vectors $(0,1,0,0,1),(1,1,1,0,1),(0,1,0,1,1),(1,1,1,1,1)$ in $V_5$ over the field of rational numbers are linearly dependent ?

If I choose the determinant method ,then do I have create a dummy row $(1,1,1,1,1)$ and form the determinant $\begin{vmatrix} 0 &1 &0 &0 &1 \\ 1 &1 &1 &0 &1 \\ 0&1 &0 &1 &1 \\ 1&1 &1 &1 &1 \\ 1&1 &1 &1 &1 \end{vmatrix}$ and the value of the determinant being $0$,the given vectors are L.D.

2.Show that the vectors $a=t^3+3t+4, b=t^3+4t+3\,\,$ are not linearly dependent?

We see that $a-b=1-t$ and hence $t=1-a+b.$ Now replacing the value of $t$ in $a=t^3+3t+4\,\,$ ,we get $a=(1-a+b)^3+3(1-a+b)+4$ and expanding the expression we get the required result.

Am I going in the right direction? Is there any alternative/better way to prove the aforementioned problems?

Thanks and regards to all.

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You haven't done the first one quite right. Any time when you have a matrix with a duplicate row, the determinant is zero. You could take four linearly independent vectors and use that method and still get that they were linearly dependent. Instead, realize that $(1, 1, 1, 1, 1) = (1, 1, 1, 0, 1) + (0, 1, 0, 1, 1) - (0, 1, 0, 0, 1)$ –  AWertheim Aug 25 '13 at 16:18
thanks a lot ,sir.Got it. –  learner Aug 25 '13 at 16:25

1 Answer 1

up vote 1 down vote accepted

Two vectors $a$ and $b$ are linearly independent iff:


Multiply each vector by a different scalar and add tem together:



This means that each of the coefficients must be equal to $0$:

$$m+n=0\\ 3m+4n=0\\ 4m+3n=0$$

The only solution to this system is $m=0$ and $n=0$.

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